Answer:
The water of the saturated clayed soil is 66.67 %.
Explanation:
Given;
mass of saturated clayed soil, = 600 g
mass of dry soil sample, = 200 g
mass of water content, = - = 600 g - 200 g = 400 g
The water content is determined as;
Therefore, the water of the saturated clayed soil is 66.67 %.
Answer:
k = 1.91 × 10^-5 N m rad^-1
Workings and Solution to this question can be viewed in the screenshot below:
Answer:
Phuong works on a research project and creates a report for her boss.
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-
a=
b=
from virial equation
And
a=
b=
Now calculate V1 and V2 at given condition
Substitute given values = 1 x 10^5 , T = 373.15 and given values of coefficients we get
Solve for V1 by iterative or alternative cubic equation solver we get
Similarly solve for state 2 at P2 = 50 bar we get
Now
a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is
a=
b=
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series